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Quantum Physics I, Lecture Note 11

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Quantum Physics I, Lecture Note 11 Lecture 11 B. Zwiebach March 17, 2016 Contents 1 The Infinite Square Well 1 2 The Finite Square Well 4 1 The Infinite Square Well In our last lecture we examined the quantum wavefunction of a particle moving in a circle. Here we introduce another instructive toy model, the infinite square well potential. This forces a particle to live on an interval of the real line, the interval conventionally chosen to be x 2 [0; a]. At the ends 0 and a of the interval there are hard walls that prevent the particle from going to x > a and x < 0. The potential is defined as follows and shown in figure 1. ( 0 ; 0 <x < a ; V (x) = (1.1) 1x ≤ 0 ; x ≥ 0 It is reasonable to assume that the wavefunction must vanish in the region where the potential is Figure 1: The infinite square well potential infinite. Classically any region where the potential exceeds the energy of the particle is forbidden. Not so in quantum mechanics. But even in quantum mechanics a particle can't be in a region of infinite potential. We will be able to justify these claims by studying the more complicated finite square well in the limit as the height of the potential goes to infinity. But for the meantime we simply state the fact: (x) = 0 for x < 0 and for x > a : (1.2) Since the wavefunction must be continuous we must have that it should vanish at x = 0 and at x = a: 1 1. (x = 0) = 0 . 2. (x = a) = 0 . These are our boundary conditions. You may wonder about the continuity of the first derivative 0(x). This derivative vanishes outside the interval and continuity would say that 0 should vanish at 0 and at a. But this is impossible. A solution of Schr¨odinger'sequation (a second order differential equation) for which both the wavefunction and its derivative vanishes at a point is identically zero! If a solution exist we must accept that 0 can have discontinuities at an infinite wall. Therefore we do not impose any boundary condition on 0. The two conditions above will suffice to find a solution. In that solution 0 is discontinuous at the endpoints. In the region x 2 [0; a] the potential vanishes and the Schr¨odingerequation takes the form d2 2mE = − ; (1.3) dx2 ~2 and as we did before, one can show that the energy E must be positive (do it!). This allows us to define, as usual, a real quantity k such that 2mE 2k2 k2 ≡ ! E = ~ : (1.4) ~2 2m The differential equation is then d2 = −k2 ; (1.5) dx2 and the general solution can be written as (x) = c1 cos kx + c2 sin kx ; (1.6) with constants c1 and c2 to be determined. For this we use our boundary conditions. The condition (x = 0) = 0 implies that c1 in Eq 1.6 must be zero. The coefficient of sin kx need not be, since this function vanishes automatically for x = 0. Therefore the solution so far reads (x) = c2 sin kx : (1.7) Note that if we demanded continuity of 0 we would have to ask for 0(x = 0) = 0 and that would make c2 equal to zero, and thus identically zero. That is not a solution. There is no particle if = 0. At this point we must impose the vanishing of at x = a. nπ c sin ka = 0 ! ka = nπ ! k = : (1.8) 2 n a Here n must be an integer and the solution would be nπx (x) = N sin ; (1.9) n a with N a normalization constant. Which integers n are acceptable here? Well, n = 0 is not acceptable, because it would make the wavefunction zero. Moreover, n and −n give the same wavefunction, up to a sign. Since the sign of a wavefunction is irrelevant, it would thus be double counting to include both positive and negative n's. We restrict ourselves to n being positive integers. 2 To solve for the coefficient, we utilize the normalization condition; every n(x) must be normalized. Z a r 2 2 nπx 2 1 2 1 = N sin dx = N 2 ·!a ! N = : (1.10) 0 a a Therefore, all in all, our solutions are: r2 nπx 2k2 2π2n2 = sin ;E = ~ n = ~ ; n = 1; 2; ··· : (1.11) n a a n 2m 2ma2 Each value of n gives a different energy, implying that in the one-dimensional infinite square well there are no degeneracies in the energy spectrum! The ground state {the lowest energy state{ corresponds to n = 1 and has nonzero energy. Figure 2: The four lowest energy eigenstates for the infinite square well potential. The nth wavefunction solution n has n − 1 nodes. The solutions are alternately symmetric and antisymmetric about the midpoint x = a. Figure 2 shows the first four solutions to the 1-d infinite square well, labeled from n = 1 to n = 4. We note a few features: 1. The ground state n = 1 has no nodes. A node is a zero of the wavefunction that is not at the ends of the domain of the wavefunction. The zeroes at x = 0 and x = a do not count as nodes. Clearly 1(x) does not vanish anywhere in the interior of [0; a] and therefore it has no nodes. It 3 Figure 3: The infinite square well shifted to the left to make it symmetric about the origin. is in fact true that any normalizable ground state of a one-dimensional potential does not have nodes. 2. The first excited state, n = 2 has one node. It is at x = a, the midpoint of the interval. The second excited state, n = 3 has two nodes. The pattern in fact continues. The n-th excited state will have n nodes. a 3. In the figure the dotted vertical line marks the interval midpoint x = 2 . We note that the ground a state is symmetric under reflection about x = 2 . The first excited state is antisymmetric, indeed a its node is at x = 2 . The second excited state is again symmetric. Symmetry and antisymmetry alternate forever. 4. The symmetry just noted is not accidental. It holds, in general for potentials V (x) that are even functions of x: V (−x) = V (x). Our potential, does not satisfy this equation, but this could have been changed easily and with no consequence. We could shift the well over so that rather a a than having V (x) = 0 from 0 ≤ x ≤ a, it extends from − 2 ≤ x ≤ 2 and then it would be symmetric about the origin x = 0 (see figure 3). We will later prove that the bound states of a one-dimensional even potential are either even or odd! Here we are just seeing an example of such result. 5. The wavefunctions n(x) with n = 1; 2;::: form a complete set that can be used to expand any function in the interval x 2 [0; a] that vanishes at the endpoints. If the function does not vanish at the endpoints, the convergence of the expansion is delicate, and physically such wavefunction would be problematic as one can verify that the expectation value of the energy is infinite. 2 The Finite Square Well We now examine the finite square well, defined as follows and shown in figure 4. ( −V ; for jx j ≤ a ; V > 0 ; V (x) = 0 0 (2.12) 0 ; for jx j ≥ a : Note that the potential energy is zero for jx j > a. The potential energy is negative and equal to −V0 in the well, because we defined V0 to be a positive number. The width of the well is 2a. Note 4 Figure 4: The finite square well potential also that we have placed the bottom of the well differently than in the case of the infinite square well. The bottom of the infinite square well was at zero potential energy. If we wanted to obtain the infinite square well as a limit of the finite square well we would have to take V0 to infinity, but care is needed to compare energies. The ones in the infinite square well are measured with respect to a bottom at zero energy. The ones in the finite square well are measure with respect to a bottom at −V0. We will be interested in bound states namely, energy eigenstates that are normalizable. For this the energy E of the states must be negative. This is readily understood. If E>0, any solutions in the region x>awhere the potential vanishes would be a plane wave, extending all the way to infinity. Such a solution would not be normalizable. The energy E is shown as a dashed line in the figure. We have −V0 <E<0 : (2.13) Note that since E is negative we have E = −|−jE j. For a bound state of energy E, the energy E~ measured with respect to the bottom of the potential is E~ = E − (−V0)=V0 −jjE j > 0 : (2.14) Those E~ are the ones that can be compared with the energies of the infinite square well in the limit as V0 !!1 1. What are the bound state solutions to the Schr¨odinger equation with this potential? We have to examine how the equation looks in the various regions where the potential is constant and then use boundary conditions to match the solutions across the points where the potential is discontinuous. We have the equation Let's examine the regions, where, for simplicity, we define A(x) by d2 2m = − (E − V (x)) = α(x) ; (2.15) dx2 ~2 where we have defined the factor α(x) that multiplies the wavefunction on the right-hand side of the Schr¨odinger equation.
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